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Georgia Southern University

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Problem 4

Answer the following questions about the functions whose derivatives are given:

\begin{equation}\begin{array}{l}{\text { a. What are the critical points of } f ?} \\ {\text { b. On what open intervals is } f \text { increasing or decreasing? }} \\ {\text { c. At what points, if any, does } f \text { assume local maximum and }} \\ \quad {\text { minimum values? }}\end{array}\end{equation}

\begin{equation}f^{\prime}(x)=(x-1)^{2}(x+2)^{2}\end{equation}

Answer

see the graph

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## Discussion

## Video Transcript

okay, Silver, given this derivative F crime of X is X minus one squared times X plus, two squared and the first thing we want to do this find the critical points. Well, it's easy to see their critical points here. I'm going to be with derivatives. Zero drift. It's never going to be under signed X equals one and X equals negative, too. And then, secondly, we want to give the open intervals where F is increasing or decreasing. So controlling number line you made it to in one we know that is going to change from being increasing that you're decreasing when asked prime changes from being positive to negative. So you want to see where f crime is positive and negative. So if I take a number less than negative too well, I mean, I guess it's not so hard to see that actually have. Crime is always greater than or equal to zero because it's the product of two things that are squared so f prime. It's always going to be increasing unless zero so to the left and negative to F crimes going be positive between aged two and one, coming positive and in greater than one. It's going to be positive. And so if we look at F, it's going to be increasing from negative infinity, the native to it's going to be increasing from negative to one and it's going to be increasing from one to infinity. So what that means is that F actually has no local extreme because it never changes from being increasing to decreasing or decreasing to increasing no local extra see.